|Version 5 (modified by goldfire, 8 months ago) (diff)|
Closed Type Families with Coincident Right-Hand Sides
In GHC 7.6, it is permitted to write something like this:
type instance F Int = Int type instance F a = a
These instances surely overlap, but in the case when they do, the right-hand sides coincide. This page explores the design space around this feature with an eye toward adding instance groups, as discussed here.
Coincident overlap within a closed type family
This section discusses a part of the "Concrete Proposal" on this page to patch a potential hole in Haskell's type system.
I (Richard) would like to be able to say this:
type family And (a :: Bool) (b :: Bool) :: Bool where And False a = False And True b = b And c False = False And d True = d And e e = e
According to the main page about closed type families, "We only match a type against an equation in a closed type family when no previous equation can unify against the type." Unfortunately, that kills us here. Consider reducing And f True. The first three branches clearly don't match. The fourth matches. But, the first two branches unify (i.e., they might apply later, depending on the instantiation for f), so we're stuck. Instead, we scan through the equations at the declaration and look to find compatible equations. Two equations are compatible if their right-hand sides match whenever their left-hand sides do. In the declaration above, all equations are compatible -- if a previous equation might apply later, we'd get the same result anyway, so we're OK. Then, during matching, we only check incompatible previous equations.
Another example is in order:
type family F a where F Int = Int F Bool = Char F a = a
The only incompatible pair of equations is F Bool and F a. Why? Let's consider this piece by piece. A use site that matches F Bool will never later match F Int, so F Int is compatible with F Bool. F Int might later match something that matches F a, but the right-hand sides coincide, so these equations are compatible. On the other hand, F Bool might also later match something that matches F a and the right-hand sides don't coincide, so these equations are not compatible. The upshot is that, if F a matches a use site, we have to make sure that F Bool cannot later apply. Otherwise, we can skip that check.
This is perhaps somewhat subtle, but it seems to be the right way to do it.
The ideas discussed here stem from posts by AntC, Andy Adams-Moran, and Richard on Richard's blog post on the subject.